post_href stringlengths 57 213 | python_solutions stringlengths 71 22.3k | slug stringlengths 3 77 | post_title stringlengths 1 100 | user stringlengths 3 29 | upvotes int64 -20 1.2k | views int64 0 60.9k | problem_title stringlengths 3 77 | number int64 1 2.48k | acceptance float64 0.14 0.91 | difficulty stringclasses 3
values | __index_level_0__ int64 0 34k |
|---|---|---|---|---|---|---|---|---|---|---|---|
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1229807/Simple-solution-in-python | class Solution:
def checkZeroOnes(self, s: str) -> bool:
count0 = count1 = max0 = max1 = 0
for c in s:
if c == '0':
count1 = 0
count0 += 1
max0 = max(max0, count0)
else:
count0 = 0
count1 += 1
... | longer-contiguous-segments-of-ones-than-zeros | Simple solution in python | souravsingpardeshi | 2 | 85 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,500 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1224894/Python3-count | class Solution:
def checkZeroOnes(self, s: str) -> bool:
ones = zeros = 0
for i in range(len(s)):
if i == 0 or s[i-1] != s[i]: cnt = 0
cnt += 1
if s[i] == "0": zeros = max(zeros, cnt)
else: ones = max(ones, cnt)
return ones > zeros | longer-contiguous-segments-of-ones-than-zeros | [Python3] count | ye15 | 2 | 61 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,501 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/2054444/Python-oneliner | class Solution:
def checkZeroOnes(self, s: str) -> bool:
return len(max(s.split('0'),key=len)) > len(max(s.split('1'),key=len)) | longer-contiguous-segments-of-ones-than-zeros | Python oneliner | StikS32 | 1 | 38 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,502 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1989058/Python-beginner-friendly-solution-using-split() | class Solution:
def checkZeroOnes(self, s: str) -> bool:
one_seg = sorted(s.split("0"), key=len)
zero_seg = sorted(s.split("1"), key=len)
return len(one_seg[-1]) > len(zero_seg[-1]) | longer-contiguous-segments-of-ones-than-zeros | Python beginner friendly solution using split() | alishak1999 | 1 | 40 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,503 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1736042/Python3-or-1LS | class Solution:
def checkZeroOnes(self, s: str) -> bool:
return len(max(s.split("0"))) > len(max(s.split("1"))) | longer-contiguous-segments-of-ones-than-zeros | Python3 | 1LS | khalidhassan3011 | 1 | 33 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,504 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1678823/fast-python-using-fact-that-string-is-binary | class Solution(object):
def checkZeroOnes(self, s):
zeros = s.split("1")
ones = s.split("0")
z = len(max(zeros, key = len))
o = len(max(ones, key = len))
return o>z
``` | longer-contiguous-segments-of-ones-than-zeros | fast python using fact that string is binary | martasapizhak | 1 | 79 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,505 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1453291/Simple-Python-O(n)-two-pointer-solution | class Solution:
def checkZeroOnes(self, s: str) -> bool:
ones_len = zeros_len = 0
left = right = 0
while right < len(s)+1:
if right == len(s) or s[right] != s[left]:
if s[left] == '0':
zeros_len = max(zeros_len, right-left)
else... | longer-contiguous-segments-of-ones-than-zeros | Simple Python O(n) two pointer solution | Charlesl0129 | 1 | 105 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,506 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1453291/Simple-Python-O(n)-two-pointer-solution | class Solution:
def checkZeroOnes(self, s: str) -> bool:
max_ones_len = max_zeros_len = 0
cur_ones_len = cur_zeros_len = 0
for d in s:
if d == '1':
cur_ones_len += 1
cur_zeros_len = 0
else:
cur_zeros_len += 1
... | longer-contiguous-segments-of-ones-than-zeros | Simple Python O(n) two pointer solution | Charlesl0129 | 1 | 105 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,507 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1245345/Python-or-Simple-Solution | class Solution:
def checkZeroOnes(self, s: str) -> bool:
zeroCount = 0
oneCount = 0
contiguousOne = 0
contiguousZero = 0
for i in s:
if i == '1':
oneCount += 1
contiguousZero = max(contiguousZero, zeroCount)
... | longer-contiguous-segments-of-ones-than-zeros | Python | Simple Solution | dee7 | 1 | 103 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,508 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1241395/Python3-one-line-solution-with-list-comprehension | class Solution:
def checkZeroOnes(self, s: str) -> bool:
return max([len(f) for f in s.split("0")]) > max([len(f) for f in s.split("1")]) | longer-contiguous-segments-of-ones-than-zeros | Python3 one line solution with list comprehension | albezx0 | 1 | 91 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,509 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1240871/python-sol-faster-than-99-less-mem-than-90 | class Solution:
def checkZeroOnes(self, s: str) -> bool:
ones = 0
zeros = 0
i = 0
while (i < len(s)):
newOne = 0
newZero = 0
while i < len(s) and s[i] == "1":
newOne += 1
i += 1
if (newOne > ones):
... | longer-contiguous-segments-of-ones-than-zeros | python sol faster than 99% , less mem than 90% | elayan | 1 | 71 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,510 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1238592/PythonPython3-Soultion-with-Explanation | class Solution:
def checkZeroOnes(self, s: str) -> bool:
cntOne,cntZero = 0,0
maxOne,maxZero = 0,0
for i in s:
if i == '0':
cntZero += 1
cntOne = 0
if maxZero < cntZero:
maxZero = cntZero
else:
... | longer-contiguous-segments-of-ones-than-zeros | Python/Python3 Soultion with Explanation | prasanthksp1009 | 1 | 67 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,511 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1227239/Python-one-liner-using-regex.-32ms-Faster-than-100 | class Solution:
def checkZeroOnes(self, s: str) -> bool:
return len(max(re.findall(r'1*', s))) > len(max(re.findall(r'0*', s))) | longer-contiguous-segments-of-ones-than-zeros | Python one-liner using regex. 32ms Faster than 100% | kingjonathan310 | 1 | 37 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,512 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1227239/Python-one-liner-using-regex.-32ms-Faster-than-100 | class Solution:
def checkZeroOnes(self, s: str) -> bool:
return len(max(s.split('0'))) > len(max(s.split('1'))) | longer-contiguous-segments-of-ones-than-zeros | Python one-liner using regex. 32ms Faster than 100% | kingjonathan310 | 1 | 37 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,513 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/2811101/python-oror-do-not-write-the-same-code-repeatly-oror-time%3A-O(n)-oror-space%3A-O(1) | class Solution:
def checkZeroOnes(self, s: str) -> bool:
maxret, lst = [0, 0], [0, 0]
for i in s:
j = ord(i) - ord("0")
lst[j], lst[1 - j] = lst[j] + 1, 0
maxret[j] = max(maxret[j], lst[j])
return maxret[0] < maxret[1] | longer-contiguous-segments-of-ones-than-zeros | python || do not write the same code repeatly || time: O(n) || space: O(1) | Aaron1991 | 0 | 3 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,514 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/2579298/Python-Generalised-One-Pass-Arbitrary-number-of-symbols | class Solution:
def checkZeroOnes(self, s: str, symbol1="1", symbol2="0") -> bool:
# make a dict to save the counts
result = collections.defaultdict(int)
# make a counter for consecutive elements
counter = 0
# make a variable to save the old charact... | longer-contiguous-segments-of-ones-than-zeros | [Python] - Generalised One-Pass - Arbitrary number of symbols | Lucew | 0 | 12 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,515 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/2384668/Python-or100-Faster-or-Time-and-Space-complexity-O(N)-O(1)-or-Longer-Contiguous-Segments | class Solution:
def checkZeroOnes(self, s: str) -> bool:
max0, max1, c0, c1 = 0, 0, 0, 0
for i in range(len(s)):
if(s[i]=='1'):
c1 += 1
c0 = 0
max1= max (max1, c1)
else:
c0 += 1
c1 = 0
... | longer-contiguous-segments-of-ones-than-zeros | Python |100% Faster | Time & Space complexity O(N) ,O(1) | Longer Contiguous Segments | NITIN_DS | 0 | 15 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,516 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/2175444/Simplest-Approach-oror-Self-Explanatory-Solution | class Solution:
def checkZeroOnes(self, s: str) -> bool:
maxConsOnes = 0
maxConsZeros = 0
ones = 0
zeros = 0
length = len(s)
for i in range(length):
if s[i] == "1":
ones += 1
else:
maxCo... | longer-contiguous-segments-of-ones-than-zeros | Simplest Approach || Self Explanatory Solution | Vaibhav7860 | 0 | 33 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,517 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/2024785/Python-Clean-and-Simple! | class Solution:
def checkZeroOnes(self, s):
count0 = max0 = count1 = max1 = 0
for d in s:
if d == '0':
count0 += 1
max0 = max(count0, max0)
count1 = 0
elif d == '1':
count1 += 1
max1 = ma... | longer-contiguous-segments-of-ones-than-zeros | Python - Clean and Simple! | domthedeveloper | 0 | 48 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,518 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/2024785/Python-Clean-and-Simple! | class Solution:
def checkZeroOnes(self, s):
max0 = max(len(seg) for seg in s.split('1'))
max1 = max(len(seg) for seg in s.split('0'))
return max1 > max0 | longer-contiguous-segments-of-ones-than-zeros | Python - Clean and Simple! | domthedeveloper | 0 | 48 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,519 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1868808/Python-dollarolution | class Solution:
def checkZeroOnes(self, s: str) -> bool:
o, z, m = s.split('0'), s.split('1'), 0
for i in o:
m = max(m,len(i))
for i in z:
if m < len(i)+1:
return False
return True | longer-contiguous-segments-of-ones-than-zeros | Python $olution | AakRay | 0 | 40 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,520 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1857540/Easiest-Python3-Solution-oror-100-Faster-oror-No-Trickey-oror-100-Easy-to-Understand | class Solution:
def checkZeroOnes(self, s: str) -> bool:
ones=0
zeros=0
res_zeros=0
res_ones=0
for i in range(len(s)):
if s[i]=="1":
ones=ones+1
res_ones=max(res_ones,ones)
else:
temp=ones
... | longer-contiguous-segments-of-ones-than-zeros | Easiest Python3 Solution || 100% Faster || No Trickey || 100% Easy to Understand | RatnaPriya | 0 | 51 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,521 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1663026/Simple-Python-solution-faster-than-99.01-of-submissions | class Solution:
def checkZeroOnes(self, s: str) -> bool:
maxOne, maxZero, currentZeroSum, currentOneSum = 0,0,0,0
for i in s:
# currentOneSum = 0
# currentZeroSum = 0
if i == "1":
currentZeroSum = 0
currentOneSum += 1
el... | longer-contiguous-segments-of-ones-than-zeros | Simple Python solution faster than 99.01% of submissions | Osereme | 0 | 54 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,522 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1415744/Python3-32ms-Memory-Less-Than-98.42 | class Solution:
def checkZeroOnes(self, s: str) -> bool:
ones, zeros, t1, t2 = 0, 0, 0, 0
for i in s:
if i == '1':
t1 += 1
zeros = max(zeros, t2)
t2 = 0
else:
t2 += 1
ones = max(... | longer-contiguous-segments-of-ones-than-zeros | Python3 32ms Memory Less Than 98.42% | Hejita | 0 | 42 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,523 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1252525/Runtime%3A-36-ms-or-Memory-Usage%3A-14.4-MB-or-beats-42.74-of-Python3-submissions | class Solution:
def checkZeroOnes(self, s: str) -> bool:
l = len(s)
if l == 1:
return bool(int(s))
one = 0
zero = 0
for i in range(l):
if '0' * i in s:
if i > zero:
zero = i
if '1' * i in s:
... | longer-contiguous-segments-of-ones-than-zeros | Runtime: 36 ms | Memory Usage: 14.4 MB | beats 42.74 % of Python3 submissions | s13rw81 | 0 | 53 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,524 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1230174/Python3-simple-solution-faster-than-95-users | class Solution:
def checkZeroOnes(self, s: str) -> bool:
a,b = 0,0
for i in range(len(s)+1):
if i == len(s) or s[i] == '0':
a = max(a,b)
b = 0
else:
b += 1
return not '0'*a in s | longer-contiguous-segments-of-ones-than-zeros | Python3 simple solution faster than 95% users | EklavyaJoshi | 0 | 29 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,525 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1226957/Easy-Python-Solution-with-Comments | class Solution:
def checkZeroOnes(self, s: str) -> bool:
from collections import Counter
# count how many 0s and 1s in the string
d1 = Counter(s)
if d1['1'] == 0:
return False
else:
#from the maximum continguous 1s
for i in range(d1['1'], 0, -1):
... | longer-contiguous-segments-of-ones-than-zeros | Easy Python Solution with Comments | pasca15 | 0 | 18 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,526 |
https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/discuss/1642204/Python-Innovative-Solution | class Solution:
def checkZeroOnes(self, s: str) -> bool:
numOnes = s.count('1')
numZeros = s.count('0')
while True:
if '1'*numOnes in s:
break
else:
numOnes -= 1
while True:
if '0'*numZeros... | longer-contiguous-segments-of-ones-than-zeros | Python Innovative Solution | ElyasGoli | -1 | 30 | longer contiguous segments of ones than zeros | 1,869 | 0.603 | Easy | 26,527 |
https://leetcode.com/problems/minimum-speed-to-arrive-on-time/discuss/1225585/Python3-Concise-binary-search-code-with-comments | class Solution:
def minSpeedOnTime(self, dist: List[int], hour: float) -> int:
# the speed upper is either the longest train ride: max(dist),
# or the last train ride divide by 0.01: ceil(dist[-1] / 0.01).
# notice: "hour will have at most two digits after the decimal point"
upper = max(max(dist), cei... | minimum-speed-to-arrive-on-time | [Python3] Concise binary search code with comments | macroway | 4 | 322 | minimum speed to arrive on time | 1,870 | 0.374 | Medium | 26,528 |
https://leetcode.com/problems/minimum-speed-to-arrive-on-time/discuss/1592449/Python-Binary-Search-with-Explanation | class Solution:
def minSpeedOnTime(self, dist: List[int], hour: float) -> int:
# helper returns required time to reach the office given a speed
def getRequiredTime(speed):
time = sum([ceil(d/speed) for d in dist[:-1]])
time += dist[-1]/speed
return time
... | minimum-speed-to-arrive-on-time | [Python] Binary Search with Explanation | artod | 3 | 152 | minimum speed to arrive on time | 1,870 | 0.374 | Medium | 26,529 |
https://leetcode.com/problems/minimum-speed-to-arrive-on-time/discuss/1225281/python3-binary-Search-Solution-(Explained) | class Solution:
def minSpeedOnTime(self, dist: List[int], hour: float) -> int:
sum_ = sum(dist)
self.ans = float("inf")
def can_reach_office(speed):
time = sum([math.ceil(dist[i]/speed) for i in range(len(dist)-1)])
return (time+(dist[-1]/speed)) <= hour
... | minimum-speed-to-arrive-on-time | python3 binary Search Solution (Explained) | swap2001 | 2 | 77 | minimum speed to arrive on time | 1,870 | 0.374 | Medium | 26,530 |
https://leetcode.com/problems/minimum-speed-to-arrive-on-time/discuss/1224905/Python3-binary-search | class Solution:
def minSpeedOnTime(self, dist: List[int], hour: float) -> int:
lo, hi = 1, 10_000_001
while lo < hi:
mid = lo + hi >> 1
tm = sum((dist[i]+mid-1)//mid for i in range(0, len(dist)-1)) + dist[-1]/mid
if tm <= hour: hi = mid
else: lo = mid... | minimum-speed-to-arrive-on-time | [Python3] binary search | ye15 | 2 | 88 | minimum speed to arrive on time | 1,870 | 0.374 | Medium | 26,531 |
https://leetcode.com/problems/minimum-speed-to-arrive-on-time/discuss/2133912/C%2B%2B-Java-Python-Solution%3A-O(NlogN)-Time-Complexity | class Solution:
def check(self, arr, m):
sum = 0
for i,a in enumerate(arr):
if(math.ceil(a/m)==math.floor(a/m) or i==len(arr)-1):
sum += a/m
else:
sum += math.ceil(a/m)
return sum
def minSpeedOnTime(self, dist: List[in... | minimum-speed-to-arrive-on-time | C++ / Java / Python Solution: O(NlogN) Time Complexity | deleted_user | 1 | 62 | minimum speed to arrive on time | 1,870 | 0.374 | Medium | 26,532 |
https://leetcode.com/problems/minimum-speed-to-arrive-on-time/discuss/2133912/C%2B%2B-Java-Python-Solution%3A-O(NlogN)-Time-Complexity | class Solution:
def minSpeedOnTime(self, dist: List[int], hour: float) -> int:
l,r = 1, 10000000
ok = False
while l<=r:
mid = (l+r)//2
need = dist[-1]/mid
sm = 0
for i in range(len(dist)-1):
sm += (dist[i]+mid-1)//mid
... | minimum-speed-to-arrive-on-time | C++ / Java / Python Solution: O(NlogN) Time Complexity | deleted_user | 1 | 62 | minimum speed to arrive on time | 1,870 | 0.374 | Medium | 26,533 |
https://leetcode.com/problems/minimum-speed-to-arrive-on-time/discuss/2348640/Python-3-solution-using-Binary-Search | class Solution:
def minSpeedOnTime(self, dist: List[int], hour: float) -> int:
if hour <= len(dist)-1:
return -1
l=1
'''Here r is set to 10**7 instead of max(r) only to satisfy the edge testcases'''
r=10**7
while(l<r):
mid= l+ (r-l)//2
... | minimum-speed-to-arrive-on-time | Python 3 solution using Binary Search | SathvikPurushotham | 0 | 26 | minimum speed to arrive on time | 1,870 | 0.374 | Medium | 26,534 |
https://leetcode.com/problems/minimum-speed-to-arrive-on-time/discuss/1918980/Python-ka-Solution | class Solution:
def canReachWithinTime(self,dist,hour,speed):
curr_hr = 0
for i in range(len(dist)):
if i==len(dist)-1:
curr_hr+=dist[i]/speed
else:
curr_hr += math.ceil(dist[i]/speed)
return curr_hr<=hour
def minSpeedOnTime(se... | minimum-speed-to-arrive-on-time | Python ka Solution | Brillianttyagi | 0 | 45 | minimum speed to arrive on time | 1,870 | 0.374 | Medium | 26,535 |
https://leetcode.com/problems/minimum-speed-to-arrive-on-time/discuss/1728802/Python-Binary-Search-Solution | class Solution:
def canReachWithinTime(self,dist,hour,speed):
curr_hr=0
for i in range(0,len(dist)):
#if we are on last train , we need not get the ceil as there is no train ahead which will depart after integer hr, so to handle edge cases of floating point , take float only.
if i==le... | minimum-speed-to-arrive-on-time | Python Binary Search Solution | aryanagrawal2310 | 0 | 69 | minimum speed to arrive on time | 1,870 | 0.374 | Medium | 26,536 |
https://leetcode.com/problems/minimum-speed-to-arrive-on-time/discuss/1535696/Python-O(1)-time-O(1)-space-binary-search-solution | class Solution:
def minSpeedOnTime(self, dist: List[int], hour: float) -> int:
if hour <= len(dist)-1:
return -1
def condition(x):
cnt = 0
n = len(dist)
for i in range(n):
if i < n-1:
f = dist[i]/x
... | minimum-speed-to-arrive-on-time | Python O(1) time, O(1) space binary search solution | byuns9334 | 0 | 110 | minimum speed to arrive on time | 1,870 | 0.374 | Medium | 26,537 |
https://leetcode.com/problems/minimum-speed-to-arrive-on-time/discuss/1359494/Binary-search-97-speed | class Solution:
def minSpeedOnTime(self, dist: List[int], hour: float) -> int:
len_dist = len(dist)
if sum(dist) <= hour:
return 1
elif len_dist - 1 > hour:
return -1
elif len_dist - 1 == int(hour):
if int(hour) == hour:
return -1
... | minimum-speed-to-arrive-on-time | Binary search, 97% speed | EvgenySH | 0 | 103 | minimum speed to arrive on time | 1,870 | 0.374 | Medium | 26,538 |
https://leetcode.com/problems/jump-game-vii/discuss/1224907/Python3-prefix-sum | class Solution:
def canReach(self, s: str, minJump: int, maxJump: int) -> bool:
prefix = [0, 1]
for i in range(1, len(s)):
prefix.append(prefix[-1])
lo = max(0, i-maxJump)
hi = max(0, i-minJump+1)
if s[i] == "0" and prefix[hi] - prefix[lo] > 0: prefix... | jump-game-vii | [Python3] prefix sum | ye15 | 3 | 123 | jump game vii | 1,871 | 0.251 | Medium | 26,539 |
https://leetcode.com/problems/jump-game-vii/discuss/1224907/Python3-prefix-sum | class Solution:
def canReach(self, s: str, minJump: int, maxJump: int) -> bool:
queue, lo = [0], 0
for x in queue:
if x == len(s)-1: return True
for xx in range(max(lo+1, x+minJump), min(x+maxJump+1, len(s))):
if s[xx] == "0": queue.append(xx)
l... | jump-game-vii | [Python3] prefix sum | ye15 | 3 | 123 | jump game vii | 1,871 | 0.251 | Medium | 26,540 |
https://leetcode.com/problems/jump-game-vii/discuss/1224907/Python3-prefix-sum | class Solution:
def canReach(self, s: str, minJump: int, maxJump: int) -> bool:
lo = 0
can = [1] + [0]*(len(s)-1)
for i, x in enumerate(s):
if x == "0" and can[i]:
for ii in range(max(lo+1, i+minJump), min(i+maxJump+1, len(s))):
if s[ii] ==... | jump-game-vii | [Python3] prefix sum | ye15 | 3 | 123 | jump game vii | 1,871 | 0.251 | Medium | 26,541 |
https://leetcode.com/problems/jump-game-vii/discuss/2610505/Clean-Fast-Python3-or-2-Pointer-or-O(n)-Time-and-Space | class Solution:
def canReach(self, s: str, minJump: int, maxJump: int) -> bool:
if s[-1] == '1':
return False
n, end = len(s), minJump
reach = [True] + [False] * (n - 1)
for i in range(n):
if reach[i]:
start, end = max(i + minJump, end), min(i ... | jump-game-vii | Clean, Fast Python3 | 2 Pointer | O(n) Time & Space | ryangrayson | 2 | 57 | jump game vii | 1,871 | 0.251 | Medium | 26,542 |
https://leetcode.com/problems/jump-game-vii/discuss/1775138/Python-easy-to-read-and-understand-or-DFSBFS | class Solution:
def dfs(self, s, i):
if i < 0 or i >= len(s) or s[i] != '0':
return
if i == len(s)-1:
self.ans = True
return
for jump in range(self.low, self.high+1):
self.dfs(s, i+jump)
def canReach(self, s: str, minJump: int, max... | jump-game-vii | Python easy to read and understand | DFS,BFS | sanial2001 | 2 | 206 | jump game vii | 1,871 | 0.251 | Medium | 26,543 |
https://leetcode.com/problems/jump-game-vii/discuss/1775138/Python-easy-to-read-and-understand-or-DFSBFS | class Solution:
def canReach(self, s: str, minJump: int, maxJump: int) -> bool:
q = [0]
far = 0
while q:
ind = q.pop(0)
if ind == len(s)-1:
return True
low = max(far+1, ind+minJump)
high = min(len(s)-1, ind+maxJump)
... | jump-game-vii | Python easy to read and understand | DFS,BFS | sanial2001 | 2 | 206 | jump game vii | 1,871 | 0.251 | Medium | 26,544 |
https://leetcode.com/problems/jump-game-vii/discuss/2057612/BFS-Python-Solution-oror-80-Faster-oror-Memory-less-than-70 | class Solution:
def canReach(self, s: str, minJump: int, maxJump: int) -> bool:
q=deque([0]) ; farthest=0
while q:
i=q.popleft()
for j in range(max(i+minJump,farthest+1), min(i+maxJump+1,len(s))):
if s[j]=='0':
q.append(j)
... | jump-game-vii | BFS Python Solution || 80% Faster || Memory less than 70% | Taha-C | 1 | 116 | jump game vii | 1,871 | 0.251 | Medium | 26,545 |
https://leetcode.com/problems/jump-game-vii/discuss/1224763/Sliding-Window-with-DP-oror-Well-Explained-oror-94-faster | class Solution:
def canReach(self, s: str, minJump: int, maxJump: int) -> bool:
if s[-1] == "1": return False
n = len(s)
dp = [False]*n
dp[-1] = True
tc = 0 #number of Trues between i+minJum, i+maxJump
for i in range(n-2, -1, -1):
# Subtracting number of True from last of the sliding window
... | jump-game-vii | 🐍 {Sliding Window with DP} || Well-Explained || 94% faster | abhi9Rai | 1 | 171 | jump game vii | 1,871 | 0.251 | Medium | 26,546 |
https://leetcode.com/problems/jump-game-vii/discuss/2776661/Python-O(n)-Time-complexity-O(1)-Space-complexity-(with-explanation) | class Solution:
def canReach(self, s: str, minJump: int, maxJump: int) -> bool:
if s[-1] == "1": # If last one equal to "1", then just return False.
return False
if minJump <= len(s)-1 <= maxJump: # If the first jump can reach the end point, then return True.
return True
... | jump-game-vii | Python O(n) Time complexity, O(1) Space complexity (with explanation) | child70370636 | 0 | 9 | jump game vii | 1,871 | 0.251 | Medium | 26,547 |
https://leetcode.com/problems/jump-game-vii/discuss/2258360/Simple-Python-Solution-with-Explanation | class Solution:
def canReach(self, s: str, minJump: int, maxJump: int) -> bool:
if s[-1] != '0':
return False
n = len(s)
dp = [False] * n
dp[0] = True
i, idx_reached = 0, minJump
while i < n and not dp[-1]:
if dp[i] == True:
j =... | jump-game-vii | Simple Python Solution with Explanation | atiq1589 | 0 | 71 | jump game vii | 1,871 | 0.251 | Medium | 26,548 |
https://leetcode.com/problems/jump-game-vii/discuss/1359910/Simulate-jumps-98-speed | class Solution:
def canReach(self, s: str, minJump: int, maxJump: int) -> bool:
if s[-1] == "1":
return False
if minJump == 1:
s = "0" + s.lstrip("0")
if s == "0":
return True
landing_idx = set(i for i, c in enumerate(s) if c == "0")
... | jump-game-vii | Simulate jumps, 98% speed | EvgenySH | 0 | 188 | jump game vii | 1,871 | 0.251 | Medium | 26,549 |
https://leetcode.com/problems/jump-game-vii/discuss/1242659/Clean-Python-BFS-approach | class Solution:
def canReach(self, s: str, minJump: int, maxJump: int) -> bool:
n=len(s)
queue=[0]
visited=set([0])
mx=0
while queue:
node=queue.pop(0)
if node==n-1:
return True
... | jump-game-vii | Clean Python ,BFS approach | jaipoo | 0 | 149 | jump game vii | 1,871 | 0.251 | Medium | 26,550 |
https://leetcode.com/problems/jump-game-vii/discuss/1224821/dp-and-sliding-window | class Solution:
def canReach(self, s: str, minJ: int, maxJ: int) -> bool:
n=len(s)
dp=[1]+[0]*(n-1)
ct=1
for i in range(minJ,n):
if ct and s[i]=='0':
dp[i]=1
ct+=dp[i+1-minJ]-dp[i-maxJ]*(i>=maxJ)
return dp[-1]==1 | jump-game-vii | dp and sliding window | leetcode_dafu | 0 | 99 | jump game vii | 1,871 | 0.251 | Medium | 26,551 |
https://leetcode.com/problems/stone-game-viii/discuss/1224872/Top-Down-and-Bottom-Up | class Solution:
def stoneGameVIII(self, s: List[int]) -> int:
s, res = list(accumulate(s)), 0
for i in range(len(s) - 1, 0, -1):
res = s[i] if i == len(s) - 1 else max(res, s[i] - res)
return res | stone-game-viii | Top-Down and Bottom-Up | votrubac | 49 | 2,700 | stone game viii | 1,872 | 0.524 | Hard | 26,552 |
https://leetcode.com/problems/stone-game-viii/discuss/1505616/6-lines-python-O(n)-time-and-O(1)-space-fater-than-99 | class Solution:
def stoneGameVIII(self, stones: List[int]) -> int:
s = sum(stones)
dp = s
for i in range(len(stones)-2, 0, -1):
s -= stones[i+1]
dp = max(dp, s - dp)
return dp | stone-game-viii | 6 lines python - O(n) time and O(1) space, fater than 99% | pureme | 1 | 170 | stone game viii | 1,872 | 0.524 | Hard | 26,553 |
https://leetcode.com/problems/stone-game-viii/discuss/1388024/Python3-Bottom-Up-O(N)-Solution-or-Very-Short | class Solution:
def stoneGameVIII(self, stones: List[int]) -> int:
sums, memory, n = [0], {}, len(stones)
for s in stones:
sums.append(s + sums[-1])
memory, biggest = [sums[-1]] * n, sums[-1]
for start in range(n - 2, -1, -1):
memory[start] = max(big... | stone-game-viii | Python3 Bottom Up O(N) Solution | Very Short | yiseboge | 1 | 137 | stone game viii | 1,872 | 0.524 | Hard | 26,554 |
https://leetcode.com/problems/stone-game-viii/discuss/1225850/Python3-top-down-dp | class Solution:
def stoneGameVIII(self, stones: List[int]) -> int:
prefix = [0]
for x in stones: prefix.append(prefix[-1] + x)
@cache
def fn(i):
"""Return max score difference."""
if i+1 == len(stones): return prefix[-1]
return max(fn(i+1... | stone-game-viii | [Python3] top-down dp | ye15 | 1 | 105 | stone game viii | 1,872 | 0.524 | Hard | 26,555 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1356591/Easy-Python-Solution(98.80) | class Solution:
def countGoodSubstrings(self, s: str) -> int:
count=0
for i in range(len(s)-2):
if(s[i]!=s[i+1] and s[i]!=s[i+2] and s[i+1]!=s[i+2]):
count+=1
return count | substrings-of-size-three-with-distinct-characters | Easy Python Solution(98.80%) | Sneh17029 | 13 | 878 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,556 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1581918/Python-Sliding-Window-Solution | class Solution:
def countGoodSubstrings(self, s: str) -> int:
k = 3
if k > len(s):
return 0
letter_frequency = {}
count, windowStart = 0, 0
for windowEnd in range(len(s)):
if s[windowEnd] not in letter_frequency:
letter_frequen... | substrings-of-size-three-with-distinct-characters | [Python] - Sliding Window Solution | TheBatmanNinja | 5 | 567 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,557 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1643675/Succinct-Python-solution-O(n) | class Solution:
def countGoodSubstrings(self, s: str) -> int:
def match(t):
return len(set(t)) == 3
return sum(map(match, zip(s, s[1:], s[2:]))) | substrings-of-size-three-with-distinct-characters | Succinct Python solution, O(n) | emwalker | 4 | 229 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,558 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1366864/Easy-to-understand-Python3-faster-than-99.83-of-Python3 | class Solution:
def countGoodSubstrings(self, s: str) -> int:
counter = 0
for i in range(len(s)-2):
new = s[i:i+3]
if len(new) == len(set(new)):
counter = counter + 1
return counter | substrings-of-size-three-with-distinct-characters | Easy to understand Python3 - faster than 99.83% of Python3 | cjenn26 | 2 | 74 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,559 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2810964/Python-oror-Sliding-Window-oror-95.01-Faster-oror-5-Lines | class Solution:
def countGoodSubstrings(self, s: str) -> int:
c,n=0,len(s)
for i in range(n-2):
t=set(s[i:i+3])
if len(t)==3:
c+=1
return c | substrings-of-size-three-with-distinct-characters | Python || Sliding Window || 95.01% Faster || 5 Lines | DareDevil_007 | 1 | 85 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,560 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1992350/Python-Solution-%2B-One-Liner! | class Solution:
def countGoodSubstrings(self, s):
count = 0
for i in range(2, len(s)):
if len(set(s[i-2:i+1])) == 3:
count += 1
return count | substrings-of-size-three-with-distinct-characters | Python - Solution + One-Liner! | domthedeveloper | 1 | 185 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,561 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1992350/Python-Solution-%2B-One-Liner! | class Solution:
def countGoodSubstrings(self, s):
return sum(len(set(s[i-2:i+1])) == 3 for i in range(2, len(s))) | substrings-of-size-three-with-distinct-characters | Python - Solution + One-Liner! | domthedeveloper | 1 | 185 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,562 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1943515/Python-or-Concise-Solution | class Solution:
def countGoodSubstrings(self, s: str) -> int:
count = 0
currString = ""
for i in range(0, len(s) - 2):
currString = s[i:i+3]
if len(set(currString)) == len(currString):
count += 1
currString = ""
... | substrings-of-size-three-with-distinct-characters | [Python] | Concise Solution | i_architect | 1 | 92 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,563 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1891630/Python-easy-solution-faster-than-85 | class Solution:
def countGoodSubstrings(self, s: str) -> int:
count = 0
for i in range(len(s)-2):
if len(set(s[i:i+3])) == len(s[i:i+3]):
count += 1
return count | substrings-of-size-three-with-distinct-characters | Python easy solution faster than 85% | alishak1999 | 1 | 70 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,564 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1803891/1-Line-Python-Solution-oror-82-Faster-oror-Memory-less-than-80 | class Solution:
def countGoodSubstrings(self, s: str) -> int:
return sum([len(set(s[i:i+3]))==len(s[i:i+3]) for i in range(len(s)-2)]) | substrings-of-size-three-with-distinct-characters | 1-Line Python Solution || 82% Faster || Memory less than 80% | Taha-C | 1 | 47 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,565 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1239037/Python3-1-line | class Solution:
def countGoodSubstrings(self, s: str) -> int:
return sum(len(set(s[i-3:i])) == 3 for i in range(3, len(s)+1)) | substrings-of-size-three-with-distinct-characters | [Python3] 1-line | ye15 | 1 | 45 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,566 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1239037/Python3-1-line | class Solution:
def countGoodSubstrings(self, s: str) -> int:
ans = 0
for i in range(3, len(s)+1):
if len(set(s[i-3:i])) == 3: ans += 1
return ans | substrings-of-size-three-with-distinct-characters | [Python3] 1-line | ye15 | 1 | 45 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,567 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1239037/Python3-1-line | class Solution:
def countGoodSubstrings(self, s: str) -> int:
ans = 0
freq = {}
for i, x in enumerate(s):
freq[x] = 1 + freq.get(x, 0)
if i >= 3:
freq[s[i-3]] -= 1
if freq[s[i-3]] == 0: freq.pop(s[i-3])
if len(freq) == 3: ... | substrings-of-size-three-with-distinct-characters | [Python3] 1-line | ye15 | 1 | 45 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,568 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1238779/Simple-Python3-Straight-Forward-Solution | class Solution:
def countGoodSubstrings(self, s: str) -> int:
cnt = 0
for i in range(2, len(s)):
if s[i] != s[i-1] and s[i] != s[i-2] and s[i-1] != s[i-2]:
cnt += 1
return cnt | substrings-of-size-three-with-distinct-characters | Simple Python3 Straight Forward Solution | pandeyrishabh40 | 1 | 50 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,569 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2837251/Sliding-Window-%2B-Set-Easy-to-Understand. | class Solution:
def countGoodSubstrings(self, s: str) -> int:
good_strings = 0
if len(s) <= 2:
return 0
fp = 0
lp = 3
for fp in range(len(s)-2):
small_string = s[fp:lp]
print(small_string)
if len(set(small_string)) == 3:
... | substrings-of-size-three-with-distinct-characters | Sliding Window + Set, Easy to Understand. | mephiticfire | 0 | 1 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,570 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2821800/Simple-Sliding-window-technique | class Solution:
def countGoodSubstrings(self, s: str) -> int:
if len(s)<3:
return 0
i=0
j=0
subString = ""
count =0
while(j<len(s)):
subString=subString+s[j]
print(subString)
if (j-i+1<3):
j+=1
... | substrings-of-size-three-with-distinct-characters | Simple Sliding window technique | rhi_1 | 0 | 3 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,571 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2813591/Beginner-friendly-with-explanation-(Sliding-Window) | class Solution:
def countGoodSubstrings(self, s: str) -> int:
i = 0
countOfGoodSubStr = 0
k = 3
j = k
while j <= len(s):
x = s[i:j]
if len(set(x)) == k:
countOfGoodSubStr += 1
i += 1
j += 1
return countO... | substrings-of-size-three-with-distinct-characters | Beginner friendly with explanation (Sliding Window) | karanvirsagar98 | 0 | 2 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,572 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2805778/Python-Fast-solution-using-a-set | class Solution:
def countGoodSubstrings(self, s: str) -> int:
ans = 0
for i in range(len(s) - 2):
ans += len(set(s[i:i + 3])) == 3
return ans | substrings-of-size-three-with-distinct-characters | [Python] Fast solution using a set | Mark_computer | 0 | 1 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,573 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2805778/Python-Fast-solution-using-a-set | class Solution:
def countGoodSubstrings(self, s: str) -> int:
return sum(len(set(s[i:i + 3])) == 3 for i in range(len(s) - 2)) | substrings-of-size-three-with-distinct-characters | [Python] Fast solution using a set | Mark_computer | 0 | 1 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,574 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2797389/Python3-simpler-code-for-Good-substring | class Solution:
def countGoodSubstrings(self, s: str) -> int:
n = len(s)
k=3
good = 0
for i in range(n-k+1):
string = s[i:k+i]
unique = set(string) //Used to find unique elements in a string
if len(unique) == 3:
good+=1
retu... | substrings-of-size-three-with-distinct-characters | Python3 simpler code for Good substring | sukumaran1004 | 0 | 1 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,575 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2778185/Stack-Python-solution | class Solution:
def countGoodSubstrings(self, s: str) -> int:
if len(s) < 3:
return 0
stack = [s[0], s[1]]
ans = 0
for i in s[2:]:
stack.append(i)
if len(stack) == 3 and len(set(stack)) == len(stack):
ans +=1
stack.p... | substrings-of-size-three-with-distinct-characters | Stack Python solution | kruzhilkin | 0 | 2 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,576 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2719817/Sliding-window-O(1)-time-solution-for-flexible-substring-size-(k-not-always-3) | class Solution:
def countGoodSubstrings(self, s: str) -> int:
k = 3
if len(s) < k:
return 0
left, right = 0, 0
chars = {}
while right < k:
right_char = s[right]
chars[right_char] = chars[right_char] + 1 if right_char in chars else 1
... | substrings-of-size-three-with-distinct-characters | Sliding window O(1) time solution for flexible substring size (k not always 3) | kudratkhujamusaev | 0 | 5 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,577 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2712055/pyhton-simple-solution-faster | class Solution:
def countGoodSubstrings(self, s: str) -> int:
res=0
for i in range(2,len(s)):
d=s[i]+s[i-1]+s[i-2]
if(len(set(d))==len(d)):
res+=1
return res | substrings-of-size-three-with-distinct-characters | pyhton simple solution faster | Raghunath_Reddy | 0 | 2 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,578 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2505010/Simplest-python-solution. | class Solution:
def countGoodSubstrings(self, s: str) -> int:
count = 0
for i in range(len(s)-2):
if len(set([s[i],s[i+1],s[i+2]])) == 3:
count+=1
return count | substrings-of-size-three-with-distinct-characters | Simplest python solution. | EbrahimMG | 0 | 34 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,579 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2504658/Simple-Python-solution | class Solution:
def countGoodSubstrings(self, s: str) -> int:
res = 0
while(len(s) >= 3):
temp = s[:3]
if len(set(temp)) == len(temp):
res += 1
s = s[1:]
return res | substrings-of-size-three-with-distinct-characters | Simple Python solution | Wartem | 0 | 16 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,580 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2499198/python-solution-or-sliding-window | class Solution:
def countGoodSubstrings(self, s: str) -> int:
i=j=0
n=len(s)
temp=[]
ans=0
while j<n:
temp.append(s[j])
if j-i+1<3:
j+=1
elif j-i+1==3:
if len(temp)==len(set(temp)):
ans+=1... | substrings-of-size-three-with-distinct-characters | python solution | sliding window | ayushigupta2409 | 0 | 24 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,581 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2462041/Python3-Sliding-Window-Clean-Solution-beats-98 | class Solution:
def countGoodSubstrings(self, s: str) -> int:
front = back = count = 0
while back < len(s):
if back - front + 1 == 3:
if s[front] != s[front+1] and s[front+1]!= s[front+2] and s[front] != s[front+2]:
... | substrings-of-size-three-with-distinct-characters | Python3 Sliding Window Clean Solution beats 98 % | abhisheksanwal745 | 0 | 19 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,582 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2427296/Using-set-and-array-indexing-in-Python | class Solution:
def countGoodSubstrings(self, s: str) -> int:
sb = []
for i in range(len(s)):
if i + 3 > len(s):
break
sb.append(s[i: i + 3])
res = []
for w in sb:
if len(set(w)) == len(w):
res.append(w)
retu... | substrings-of-size-three-with-distinct-characters | Using set and array indexing in Python | ankurbhambri | 0 | 20 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,583 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2418613/Python-Easy-solution | class Solution:
def countGoodSubstrings(self, s: str) -> int:
count=0
for i in range(len(s)):
s1=set(s[i:i+3])
if(len(s1)==3):
count+=1
return count | substrings-of-size-three-with-distinct-characters | Python Easy solution | mrigank2303239 | 0 | 21 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,584 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2377498/Python-Solution-or-One-Liner-or-Kind-Of-Sliding-Window | class Solution:
def countGoodSubstrings(self, s: str) -> int:
return sum(len(set(s[i:i+3])) == 3 for i in range(0,len(s)-2)) | substrings-of-size-three-with-distinct-characters | Python Solution | One - Liner | Kind Of Sliding Window | Gautam_ProMax | 0 | 26 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,585 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2283733/Simple-python-solution-with-18ms-runtime | class Solution(object):
def isValid(self, temp):
thedict = {}
for i in range(len(temp)):
if temp[i] not in thedict:
thedict[temp[i]] = 0
else:
return False
return True
def countGoodSubstrings(self... | substrings-of-size-three-with-distinct-characters | Simple python solution with 18ms runtime | Gilbert770 | 0 | 25 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,586 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2184730/Python-simple-solution | class Solution:
def countGoodSubstrings(self, s: str) -> int:
ans = 0
for i in range(3,len(s)+1):
if len(set(s[i-3:i])) == 3:
ans += 1
return ans | substrings-of-size-three-with-distinct-characters | Python simple solution | StikS32 | 0 | 34 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,587 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2174846/FASTEST-SOLUTION-100-faster-than-python-solutions | class Solution:
def countGoodSubstrings(self, s: str) -> int:
window_size = 3 # static window
result = 0
if len(s) >= window_size:
for start in range(len(s) - window_size + 1):
result += (1 if (len(set(s[start:start + window_size])) == 3) else 0)
... | substrings-of-size-three-with-distinct-characters | FASTEST SOLUTION 100% faster than python solutions | notxkaran | 0 | 67 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,588 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/2079596/Basic-sliding-window | class Solution:
def countGoodSubstrings(self, s: str) -> int:
res = i = 0
while i <= len(s) - 3:
sub = set(s[i:i+3])
if len(sub) == 3:
res += 1
i += 1
return res | substrings-of-size-three-with-distinct-characters | Basic sliding window | andrewnerdimo | 0 | 104 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,589 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1954159/Python3-simple-solution | class Solution:
def countGoodSubstrings(self, s: str) -> int:
count = 0
for i in range(len(s)-2):
if s[i] != s[i+1] and s[i] != s[i+2] and s[i+1] != s[i+2]:
count += 1
return count | substrings-of-size-three-with-distinct-characters | Python3 simple solution | EklavyaJoshi | 0 | 27 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,590 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1927232/Python-3-Solution | class Solution:
def countGoodSubstrings(self, s: str) -> int:
c = 0
i = 0
j = 1
word = ""
while j < len(s):
if j - i + 1 < 3:
word += s[i]
if s[j] not in word:
word += s[j]
j += 1
else... | substrings-of-size-three-with-distinct-characters | Python 3 Solution | AprDev2011 | 0 | 47 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,591 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1868883/Python-dollarolution | class Solution:
def countGoodSubstrings(self, s: str) -> int:
count = 0
for i in range(len(s)-2):
if len(set((s[i],s[i+1],s[i+2]))) == 3:
count += 1
return count | substrings-of-size-three-with-distinct-characters | Python $olution | AakRay | 0 | 37 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,592 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1775898/Simple-and-Easy-to-understand-solution | class Solution:
def countGoodSubstrings(self, s: str) -> int:
ans1=[]
for i in range(len(s)-2):
ans=""
ans+=s[i:i+3]
ans1.append(ans)
count=0
for a in ans1:
if len(set(a)) ==3:
count+=1
return count | substrings-of-size-three-with-distinct-characters | Simple and Easy to understand solution | Buyanjargal | 0 | 49 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,593 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1621512/Python.-Easy-and-Fast.-24-36ms-runtime. | class Solution:
def countGoodSubstrings(self, s: str) -> int:
i = 0
goodStrings = 0
while i < len(s)-2:
currSet = set(s[i:i+3])
if len(currSet) == 3:
goodStrings +=1
i+=1
elif len(currSet) == 1:
i+=2... | substrings-of-size-three-with-distinct-characters | Python. Easy and Fast. 24-36ms runtime. | manassehkola | 0 | 55 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,594 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1595825/Simple-Python-Solution | class Solution:
def countGoodSubstrings(self, s: str) -> int:
ans = 0
if len(s) < 3: return ans
slow = 0
fast = slow+2
while fast < len(s):
if len(Counter(s[slow:fast+1])) == 3:
ans += 1
... | substrings-of-size-three-with-distinct-characters | Simple Python Solution | anandudit | 0 | 49 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,595 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1496372/2-solutions-in-Python-one-liner-and-sliding-window | class Solution:
def countGoodSubstrings(self, s: str) -> int:
return sum(len(set(s[i:(i + 3)])) == 3 for i in range(len(s) - 2)) | substrings-of-size-three-with-distinct-characters | 2 solutions in Python, one-liner and sliding-window | mousun224 | 0 | 129 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,596 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1496372/2-solutions-in-Python-one-liner-and-sliding-window | class Solution:
def countGoodSubstrings(self, s: str) -> int:
cnt = j = 0
seen = dict()
for i, c in enumerate(s):
if c in seen and i - seen[c] < 3:
j = seen[c] + 1
elif i - j == 2:
cnt += 1
j += 1
seen[c] = i... | substrings-of-size-three-with-distinct-characters | 2 solutions in Python, one-liner and sliding-window | mousun224 | 0 | 129 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,597 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1493872/python-one-liner-O(n)-solution-with-zip | class Solution:
def countGoodSubstrings(self, s: str) -> int:
return sum(a1 != a2 != a3 != a1 for a1, a2, a3 in zip(s, s[1:], s[2:])) | substrings-of-size-three-with-distinct-characters | [python] one-liner O(n) solution with zip | licpotis | 0 | 35 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,598 |
https://leetcode.com/problems/substrings-of-size-three-with-distinct-characters/discuss/1389313/Simple-Python-Solution | class Solution:
def countGoodSubstrings(self, s: str) -> int:
count=0
for i in range(len(s)-2):
ele=s[i:i+3]
if len(set(ele))==3:
count+=1
return count | substrings-of-size-three-with-distinct-characters | Simple Python Solution | sangam92 | 0 | 36 | substrings of size three with distinct characters | 1,876 | 0.703 | Easy | 26,599 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.